Is a DC-motor an inductive load?
Hello all
I had a discussion with a friend about loads. He said to me that anything which operated on DC is a resistive load - Never inductive. So now I ask: What about a brushed DC-motor? Isn't that an inductive load?
Now I ask you as well - Maybe you can help educating me a bit. I'm sure he's right, but why?
Your friend is completely wrong!
Every motor is an inductive load!
Let him take apart a motor and have him check if it contains any coils. I bet he'll find almost only coils.
It's the coil's magnetic force making the motor rotate.
Boncuk
I'm not a complete idiot. Some parts are still missing.
HAH!
YES it is ABSOLUTELY an inductive load. It is operated at DC, but the inductive component looks like a short at DC. Try and switch it off real quick and you'll see how inductive it really is.
He's not right. In reality everything is a little bit of everything else since nothing is ideal (resistance, capacitance, inductance). But ideally...
If a brushed motor was a resistive load, then it would fry whenever you hooked up a power supply across it because the winding resistance is so low. BUt it doesn't. A DC brushed motor accepts DC current, but inside the rotating brushes are commutating (switching the DC current back and forth across the windings) so that if you measured the voltage across the individual windings you would not see a flat constant DC voltage, but you would see a AC waveform (not necessarily the pure sinusoid AC waveform you get from the mains, more likely to be a trapzoidal waveform but AC nonetheless with no AC component).
THe winding inductance is slowing down the rise time of the current in the winding which is trying to reach to "infinite" due to the low resistance of the motor windings. But before it can reach "infinite" (a level of current that would fry the motor) the brushes commutate the current to another winding.
What it's actually doing is for every winding it is switching the voltage polarity of the winding to reverse the current flow so that it decreases to zero before it gets too high. BUt once it reaches zero it starts increasing in the opposite direction. The voltage polarity is reversed again before this gets too high (assuming the motor is running under allowable operating conditions). BUt it is doing this in a sequence so it appears as though the positive voltage across windings is jumping sequentially across the windings in one direction and the negative voltage is moving in the opposite direction across the windings taking the empty place the winding that the other polarity left behind.
If there is too much load on the motor the motor is spinning too slowly and the current polarity isn't reversed fast enough and is given too much time to rise to a level that overheats the motor. In the absolute worst case (stall) then, yeah, it appears as a resistive load since the inductance will allow the DC current flow through a single winding to eventually reach a level where the current is now limited by the resistance and thus no longer rises meaning the inductance then appears as a short.
Last edited by dknguyen; 2nd September 2008 at 11:05 PM.
A spinning motor generates a back EMF which acts to reduce current flow from the locked rotor value.
An inductance generates a voltage, e, which equals -L Δi/Δt, with L being inductance in henries, Δi being the change in current and Δt being the change in time that the change in current takes place over.
There is a negative sign because the inductor tries to maintain the circuit current through itself by forcing a voltage through the circuitry; the coil voltage polarity changes from + to -.
Neglecting the effect of parasitic coil capacitance, for a 1 h coil, for a change in i of 2 amp over a time duration of 3 mS you get a 1300 v spike.
There are several ways to reduce this spike, including a fast shunting diode or just a plain shunt resistor, so the coil driving circuitry is protected.
Last edited by Willbe; 2nd September 2008 at 11:50 PM.
Your friend likely was referring to the steady-state power drawn by a load. With steady-state DC there's no inductive or capacitive reactance to provide reactive power as there is with AC so the only load you see is resistive, even if there's an inductor or capacitor in the circuit. But when you turn the circuit on or off, you're generating an AC component (step function), in which case you will definitely see the effect of the inductance and/or capacitance.
Likewise a dc motor running with a steady load will look resistive to a steady-state DC supply (although it may add some hash noise to the line from the inductive switching through the brushes). Again the motor inductance effects will show up primarily during turning the voltage on or off.
Carl
Curmudgeon Elektroniker
How about if the motor is driving a gearbox which, for example, is pulling a spring loaded piston half the time of a cycle and with no load the other half of the time while the spring returns the piston to its start position?Originally Posted by crutschow
Wouldn't that cause a variation in the load of the motor and thus cause inductive load?
... Oh, and thank you for your answers, by the way. I'm sure they will help shed some light on our debate
hi,... Oh, and thank you for your answers, by the way. I'm sure they will help shed some light on our debate
Whether the motor is running ON or OFF load there is always a back emf.
A motor winding is only seen as resistive at the instant its starting to turn.
As the motors windings dc resistance is usually quite low the starting current is high
until the back emf balances the applied motor voltage
Last edited by ericgibbs; 3rd September 2008 at 08:16 AM.
Eric " Good enough is Perfect "
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Not all motors are inductive some are capacitive or can vary from capacitive to inductive.
